introduction to digital modulation (1) (1)

1. Introduction :
Analog modulation and digital modulation
 Both analog and digital modulation systems
use analog carriers to transport the
information signal.
 In analog modulation, the information is
also analog, whereas with digital
modulation, the information is digital
which could be computer generated data or
digitally encoded analog signals.

2. Digital modulation
• Is the transmittal of digitally modulated analog
signals between two or more points in a communications
system.
• Can be propagated through Earth’s atmosphere and
used in wireless communication system - digital radio.
• Offer several outstanding advantages over
traditional analog system.
• Ease of processing
• Ease of multiplexing
• Noise immunity

3. Cont’d...
 Applications:
 Low speed voice band data comm. modems
 High speed data transmission systems
 Digital microwave & satellite comm. systems
 PCS (personal communication systems)
telephone

4. Why digital modulation?
 The modulation of digital signals with analogue
carriers allows an improvement in signal to noise
ratio as compared to analogue modulating
schemes.

5. Important Criteria
1. High spectral efficiency
2. High power efficiency
3. Robust to multipath
4. Low cost and ease of implementation
5. Low carrier-to-co channel interference ratio
6. Low out-of-band radiation

6. Cont’d…
7. Constant or near constant envelop
8. Bandwidth Efficiency
 Ability to accommodate data within a limited
bandwidth
 Tradeoff between data rate and pulse width
9. Power Efficiency
 To preserve the fidelity of the digital message at
low power levels.
 Can increase noise immunity by increasing
signal power

7. Forms of Digital Modulation
)
2
sin(
)
( 
 
 ft
V
t
v
ASK
FSK
PSK
QAM

8. Forms of Digital Modulation
)
2
sin(
)
( 
 
 ft
V
t
v
•If the amplitude, V of the carrier is varied proportional to
the information signal, a digital modulated signal is called
Amplitude Shift Keying (ASK)
•If the frequency, f of the carrier is varied proportional to
the information signal, a digital modulated signal is called
Frequency Shift Keying (FSK)

9. Cont’d…
 If the phase, θ of the carrier is varied proportional
to the information signal, a digital modulated signal
is called Phase Shift Keying (PSK)
 If both the amplitude and the phase, θ of the carrier
are varied proportional to the information signal, a
digital modulated signal is called Quadrature
Amplitude Modulation (QAM)

10. Cont’d...

11. Block Diagram
Simplified block diagram of a digital modulation system
Transmitter Receiver

12. Cont’d…
 Precoder performs level conversion & encodes
incoming data into group of bits that modulate
an analog carrier.
 Modulated carrier filtered, amplified &
transmitted through transmission medium to Rx.
 In Rx, the incoming signals filtered, amplified &
applied to the demodulator and decoder circuits
which extracts the original source information
from modulated carrier.

13.  Information capacity, Bits & Bit Rate
 Information capacity is a measure of how much
information can be propagated through a
communication system and is a function of
bandwidth and transmission time.
 Represents the number of independent symbols that
can be carried through a system in a given unit of
time.
 Basic digital symbol is the binary digit or bit.
 Express the information capacity as a bit rate – the
number of bits transmitted during one second (bps).
Information Capacity, Bits,
Bit Rate, Baud, M-ary Encoding

14. Hartley’s Law
t
B
I 

Where
I = information capacity (bps)
B = bandwidth (Hz)
t = transmission time (s)
From the equation, Information capacity is a
linear function of bandwidth and transmission
time and directly proportional to both.

15. Shannon’s Formula
)
1
(
log
32
.
3
)
1
(
log 10
2 N
S
N
S
B
I
or
B
I 



Where
I = information capacity (bps)
B = bandwidth (Hz)
= signal to noise power ratio (unitless)
The higher S/N the better the performance and the
higher the information capacity
N
S

16. Example
By using the Shannon’s Formula, calculate
the information capacity if S/N = 30 dB and
B = 2.7 kHz.
)
1
(
log2 N
S
B
I 

kbps
I 9
.
26
)
1000
1
(
log
)
2700
(
32
.
3 2 



17. Nyquist Sampling Rate
 fs is equal or greater than 2fm
fs >= 2fm
fs = minimum Nyquist sample rate (Hz)
fm = maximum analog input frequency (Hz)

18. Example
Determine the Nyquist sample rate for
a maximum analog input frequency 7.5
kHz.
fs >= 2fm
fs >= 2(7.5kHz) = 15kHz

19. M-ary Encoding
 It is often advantageous to encode at a level higher
than binary where there are more then two
conditions possible.
 The number of bits necessary to produce a given
number of conditions is expressed mathematically
as
M
N 2
log

Where N = number of bits necessary
M = number of conditions, level or combinations
possible with N bits.
OR
N
M 2


20. Cont’d…
 Each symbol represents n bits, and has M
signal states, where M = 2N
.
 Example;
A digital signal with four possible conditions
(voltage levels, frequencies, etc) is an M-ary
system with number of possible conditions,
M=4.

21. Example
Find the number of voltage levels
which can represent an analog signal
with
a. 3 Bits
b. 8 bits
c. 12 bits
Ans: M=8,256,4096

22. Baud
 Baud refers to the rate of change of a signal on
the transmission medium after encoding and
modulation have occurred.
Where
baud = symbol rate (symbol per second)
ts = time of one signaling element or symbol
(seconds)
s
t
baud
1

ts

23. Bit rate and baud
 Bit rate refers to the rate of change of
a digital information signal, which is
usually binary. (bps or b/s )
 Baud, like bit rate, also a rate of
change but it refers to the rate of
change of a signal on transmission
medium after encoding and modulation
process.

24. Minimum BW
 Minimum Bandwidth refers to the minimum bandwidth
necessary to pass M-ary digitally modulated carriers.
 From the Nyquist formulation for channel capacity, fb
M
B
fb 2
log

Where fb= channel capacity (bps)
B = minimum Nyquist bandwidth (Hz)
M = number of discrete signal or voltage levels
Then,
N
f
M
f
B b










2
log
Where N is the number of bits encoded into each signaling element.

25. Example
Determine the minimum bandwidth and
baud necessary to pass a 10 kbps binary
signal using amplitude shift keying.
N
f
M
f
B b
b










2
log
Solution
ASK : N=1, and the minimum bandwidth are
baud
f
f
B b
b










 000
,
10
1
1
000
,
10

26. Additional note
 N=1  1 bit is
represented for a
signaling element
or symbol.
ts
N=1, gives the following equation becomes
N
f
M
f
B b










2
log
= fb , in binary system baud = bit
per second are equal.
In higher system, bps always greater than baud.

27. Bandwidth Efficiency
 Used to compare the performance of one digital
modulation technique to another.
Bη = Transmission bit rate (bps)
Minimum bandwidth (Hz)